# Matrices with the same characteristic polynomial

For all the $n \times n$ matrices, let's define an equivalent relation that two matrices are in the relation iff they have the same characteristic polynomial.

How can we characterize the matrices that share the same characteristic polynomial?

Why is the characteristic polynomial for a matrix called "characteristic", if there can be multiple matrices sharing the same characteristic polynomial? Compared to probability, in probability, we also have a characteristic function for a probability distribution, and different distributions have different characteristic functions, so a characteristic function does characterize a distribution.

Thanks.

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I think the best you can hope for is that matrices with the same characteristic polynomial have the same eigenvalues (multiplicities and all). –  Git Gud Jul 3 at 23:32
But do matrices that have the same eigenvalues and multiplicities have the same characteristic polynomial? I doubt it. For example, scaling a polynomial doesn't change its roots and multiplicities. I am not sure if there are other examples. –  Tim Jul 3 at 23:33
@Tim if two matrices have the same eigenvalues with the same (algebraic) multiplicities, then they indeed have the same characteristic polynomial –  Omnomnomnom Jul 3 at 23:34
@Tim By the fundamental theorem of algebra, the characteristic polynomial is completely determined by its roots, which are eigenvalues (multiplicities included). –  Tunococ Jul 3 at 23:39
In general it's not worth taking names too seriously. Things get named for all sorts of weird reasons. Don't worry about it too much. –  Qiaochu Yuan Jul 4 at 0:25

The characteristic polynomial $p(x) = \det(x I - A)$ of any matrix $A$ can be factored as $$p(x) = (x-\lambda_1)^{d_1}\cdots(x - \lambda_k)^{d_k}$$ where each $d_i$ is a positive integer and no two $\lambda_i \in \mathbb C$ are equal.

The $\lambda_i$ that form the roots of the characteristic polynomial are precisely the eigenvalues of our matrix, and the value $d_i$ is called the "algebraic-multiplicity" of the eigenvalue $\lambda_i$, which corresponds the dimension of the generalized eigenspace associated with $\lambda_i$.

In fact, if all $d_i$ are equal to $1$, then we can guarantee that any two matrices with this characteristic polynomial are similar.

(1)

Ultimately, a name is just a name. In a sense, it would be impossible to capture all the information of an $n \times n$ matrix is just an $n^{th}$ degree polynomial on one variable, and so it follows that there are many matrices which share a characteristic polynomial.

That being said, the characteristic polynomial gives us a lot of the information that we care about. Namely, it tells us what the eigenvalues are. Particularly, in the case that an $n \times n$ matrix has $n$ distinct eigenvalues (which, in certain arenas, is almost certain), another matrix will have the same characteristic polynomial if and only if it is similar to the first. In an important sense, similar matrices are "basically the same", i.e. "the same, up to a relabeling of some vectors", i.e. "the same, up to a change in basis".

(2)

When the multiplicities are not all one, all we can say is that the two matrices have the same "generalized eigenspaces", which in the end of the day is not much. If you really want to categorize all of your matrices into classes of similar matrices, you'll need something like Schur decomposition or Jordan normal form. Or, once the characteristic polynomial is obtained, the "Weyr characteristic" will allow you classify the matrix up to similarity.

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Thanks. (1) Why is the characteristic polynomial for a matrix called "characteristic", if there can be multiple matrices sharing the same characteristic polynomial? Compared to probability, in probability, we also have a characteristic function for a probability distribution, and different distributions have different characteristic functions, so a characteristic function does characterize a distribution. –  Tim Jul 3 at 23:49
(2)if the multiplicities are not all 1, what can we say about the matrices sharing the same characteristic polynoimal? Other than they having the same eigenvalues, eigenspaces? –  Tim Jul 3 at 23:51
They won't even necessarily have the same eigenspaces! Consider the matrices $\begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ –  DavidButlerUofA Jul 3 at 23:55
@DavidButlerUofA: Then if the multiplicities are not all 1, what can we say about the matrices sharing the same characteristic polynoimal? –  Tim Jul 4 at 0:00
I was trying to make matrices whose multiplicities weren't 1, since you asked abput matrices where they weren't 1. Can I check something? Do you think that these two matrices have different characteristic polynomial? –  DavidButlerUofA Jul 4 at 0:04